Question

A particle of mass 'm' and charge 'q', initially at rest, is accelerated by a uniform electric field 'E' through a distance 'D' and is then allowed to approach a fixed static charge 'Q' of the same sign. The distance of the closest approach of the charge q is
[ \(ε_0\) = permittivity of free space ]

• A. \(Q/4πε_0D\)
• B. \(Q/4πε_0ED\)
• C. \(Q/2πε_0D^2\)
• D. \(Q/4πε_0E\)

Hint:

Work done by the field becomes kinetic energy, which is then converted to potential energy at the closest approach.

Answer 1

The Correct Option is B

Step 1: Understand the concept
The charge gains kinetic energy equal to the work done by the field. Later, at the closest approach, all that kinetic energy has become electrostatic potential energy.

Step 2: Energy gained
The force is \(qE\) over a distance \(D\), so the kinetic energy gained is \(K = qED\).

Step 3: Energy at closest approach
At the closest distance \(r\) the speed is zero, so \(K = \dfrac{qQ}{4\pi\varepsilon_0 r}\) (the starting potential energy at large distance is zero).

Step 4: Solve
\[ qED = \frac{qQ}{4\pi\varepsilon_0 r} \Rightarrow r = \frac{Q}{4\pi\varepsilon_0ED} \]
Option (B). The charge \(q\) and the mass \(m\) cancel out.

Final Answer:
The closest approach is Q/(4 pi epsilon_0 E D). This is option (B). \[ \boxed{\text{(B) }\frac{Q}{4\pi\varepsilon_0ED}} \]